Solve Depth of Pool w/ Snell's Law

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SUMMARY

The discussion revolves around calculating the depth of a pool using Snell's Law, specifically the scenario where a person standing 3 meters away can see half of a stone at the bottom. The correct application of Snell's Law (n * sin(x) = n2 * sin(y)) reveals that the angles of incidence and refraction must be accurately determined. The final calculation indicates that the depth of the pool is approximately 4 meters, derived from the correct interpretation of angles and the geometry of the situation.

PREREQUISITES
  • Understanding of Snell's Law and its application in optics
  • Basic trigonometry, including sine and tangent functions
  • Knowledge of refractive indices (n for air and water)
  • Ability to visualize and interpret geometric relationships in a triangle
NEXT STEPS
  • Study the derivation and applications of Snell's Law in various mediums
  • Learn how to apply the Pythagorean theorem in real-world scenarios
  • Explore advanced trigonometric functions and their applications in physics
  • Practice problems involving light refraction and geometric optics
USEFUL FOR

Students in physics or engineering, educators teaching optics, and anyone interested in practical applications of trigonometry and Snell's Law in real-world scenarios.

BadatPhysicsguy
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Homework Statement


MGxXqsJ.png

A stone lies to the very edge at the bottom of a pool. The pool is filled with water to the top. The person standing three meters away from the pool is 1 meter tall and he can see exactly the half of the stone. Calculate the depth of the pool.

Homework Equations


Snell's law: n * sin(x) = n2 * sin(y)

The Attempt at a Solution


QHBwIM4.png

Okay.. Since I can see half of the stone, the angle should be 45 degrees, right? I should be able to calculate x degrees and therefore the bottom of the triangle (where he stands). With the bottom, I can calculate the bottom line of the triangle in the pool and then the depth. So I do that.

sin(45) * 1.33 = 1.00 * sin(x)
1.33 is n for water. 1 is n for air. x = 70.05 degrees.

So the inside of the triangle is approx 20 degrees. I use tan.

tan 70 = 1/adjacent => 1/(tan 70) = adjacent => 0.36397. So the bottom of that triangle is 0.36397 meters, but that doesn't make sense, it should be atleast 3 meters because he is standing that far away from the pool. What am I doing wrong?
 
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Angles of incidence and refraction in Snell's law are measured from the normal to the refracting surface.
 
Bystander said:
Angles of incidence and refraction in Snell's law are measured from the normal to the refracting surface.
Hello! Thank you for your answer. I did draw a normal perpendicular to the water's surface. But if is measured from the normal to the refracting surface aren't both angles 90 degrees then?
 
BadatPhysicsguy said:

Homework Statement


[ IMG]http://i.imgur.com/MGxXqsJ.png[/PLAIN]
A stone lies to the very edge at the bottom of a pool. The pool is filled with water to the top. The person standing three meters away from the pool is 1 meter tall and he can see exactly the half of the stone. Calculate the depth of the pool.

Homework Equations


Snell's law: n * sin(x) = n2 * sin(y)

The Attempt at a Solution


[ IMG]http://i.imgur.com/QHBwIM4.png[/PLAIN]
Okay.. Since I can see half of the stone, the angle should be 45 degrees, right? I should be able to calculate x degrees and therefore the bottom of the triangle (where he stands). With the bottom, I can calculate the bottom line of the triangle in the pool and then the depth. So I do that.

sin(45) * 1.33 = 1.00 * sin(x)
1.33 is n for water. 1 is n for air. x = 70.05 degrees.

So the inside of the triangle is approx 20 degrees. I use tan.

tan 70 = 1/adjacent => 1/(tan 70) = adjacent => 0.36397. So the bottom of that triangle is 0.36397 meters, but that doesn't make sense, it should be at least 3 meters because he is standing that far away from the pool. What am I doing wrong?
Seeing half the stone has nothing to do with a 45° angle.

Basically, a ray from the center of the stone should just graze the edge of the pool as it exits the water, then continue on to the eye.
 

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SammyS said:
Seeing half the stone has nothing to do with a 45° angle.

Basically, a ray from the center of the stone should just graze the edge of the pool as it exits the water, then continue on to the eye.

Hello and thank you for your answer.

But if the ray comes from the very center of the stone and it is truly in the edge of the pool, shouldn't it be "like" 45 degrees from the pool's wall down to the center of the stone?

And if the ray just grazes the edge of the pool, I then get this:
QeeP5gu.png


which would make by original attempt not work. Basically the only thing I need now is either x or y (degrees) and with it I can calculate the side (depth). But using Snell's law, now I only know the n's, not the angles. So I have two unknowns. How should I proceed?
 
BadatPhysicsguy said:
Hello and thank you for your answer.

But if the ray comes from the very center of the stone and it is truly in the edge of the pool, shouldn't it be "like" 45 degrees from the pool's wall down to the center of the stone?

And if the ray just grazes the edge of the pool, I then get this:
QeeP5gu.png


which would make by original attempt not work. Basically the only thing I need now is either x or y (degrees) and with it I can calculate the side (depth). But using Snell's law, now I only know the n's, not the angles. So I have two unknowns. How should I proceed?
Do you know trigonometry?

You can get sin(z) by using the Pythagorean theorem. Otherwise, use a scientific calculator to find the angle z

Then use Snell's Law , etc.
 
SammyS said:
Do you know trigonometry?

You can get sin(z) by using the Pythagorean theorem. Otherwise, use a scientific calculator to find the angle z

Then use Snell's Law , etc.
Hello, yes I forgot that! I calculated z degrees and then got x to be 45 degrees (making y also 45). So the depth is the same as the base of the triangle, approx 4 meters.
 
BadatPhysicsguy said:
Hello, yes I forgot that! I calculated z degrees and then got x to be 45 degrees (making y also 45). So the depth is the same as the base of the triangle, approx 4 meters.
Yes, approximately 45° , so approximately 4 meters deep.
 

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